Solve. $$ x^{2}+3 x-23 \geq x-8 $$

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Solve the equation $$ x^{2}+3x-23 \geq x-8 $$.
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
  $$x^{2}+3x-23\geq x-8$$
- step1: Move the expression to the left side:
  $$x^{2}+3x-23-\left(x-8ight)\geq 0$$
- step2: Calculate:
  $$x^{2}+2x-15\geq 0$$
- step3: Rewrite the expression:
  $$x^{2}+2x-15=0$$
- step4: Factor the expression:
  $$\left(x-3ight)\left(x+5ight)=0$$
- step5: Separate into possible cases:
  $$\begin{align}&x-3=0\&x+5=0\end{align}$$
- step6: Solve the equation:
  $$\begin{align}&x=3\&x=-5\end{align}$$
- step7: Determine the test intervals:
  $$\begin{align}&x<-5\&-53\end{align}$$
- step8: Choose a value:
  $$\begin{align}&x_{1}=-6\&x_{2}=-1\&x_{3}=4\end{align}$$
- step9: Test the chosen value:
  $$\begin{align}&x<-5	extrm{ }	extrm{is the solution}\&-53	extrm{ }	extrm{is the solution}\end{align}$$
- step10: Include the critical value:
  $$\begin{align}&x\leq -5	extrm{ }	extrm{is the solution}\&x\geq 3	extrm{ }	extrm{is the solution}\end{align}$$
- step11: The final solution is $$x \in \left(-\infty,-5ight]\cup \left[3,+\inftyight):$$
  $$x \in \left(-\infty,-5ight]\cup \left[3,+\inftyight)$$
The solution to the inequality $$x^{2}+3x-23 \geq x-8$$ is $$x \in (-\infty,-5) \cup [3,+\infty)$$.
The solution is $$ x \leq -5 $$ or $$ x \geq 3 $$.

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